ANALYSIS I

13 Power Series

13.1 Definition

Let (an) be a real or complex series. n The series generated by the sequences (anz ) as z varies are called the power series generated by (an). n We usually just speak of ‘the power series (anz ). Note. These are the most important series of all! (Taylor, Maclaurin, etc, etc.)

Note. We will prove our theorems for real (an) but almost all apply equally to real and complex sequences.

13.2 Examples

P n convergent if |x| < 1 (i) an = 1 : x : Geometric Series : divergent if |x| > 1 P n (ii) an = 1/n!: x /n! : Exponential Series : convergent for all x convergent if |x| < 1 (iii) a = 1/n : P∞ xn/n : Logarithmic Series : n n/1 divergent if |x| > 1

n  a2n = (−1) /2n! P x2n(−1)n (iv) : 2n! : Cosine Series : convergent for all x a2n+1 = 0  a2n = 0 P x2n+1(−1)n (v) n : (2n+1)! : Sine Series : convergent for all x a2n+1 = (−1) /(2n + 1)!

13.3 The Circle of Convergence P n P n Theorem. If anx0 is convergent, and |x| < |x0| then anx is absolutely convergent.

x P n n Proof. Chose p such that < p < 1. As anx is convergent, anx → 0, so there exist x0 0 0 N such that

n n > N =⇒ |anx0 | < 1 Then n n n x n n > N =⇒ |anx | 6 |anx0 | 6 p x0 N+1 N+2 P n Then by Comparison Test with (1, 1, 1,..., 1, p , p ,... ), |anx | is convergent.

1 13.4 Definition of Radius of Convergence P n P n Suppose {|x0| : anx0 convergent} is bounded; then let R := sup{|x0| : anx0 convergent}. If the set is not bounded, we write “R = ∞”. We call this the Radius of Convergence. Why? Well, see below. Example. With the same examples as above we get R = 1, ∞, 1, ∞, ∞.

13.5 Why it is the ‘radius’ P n Theorem. Let anx have radius of convergence R. Then P n (i) |x| < R then anx is absolutely convergent P n (ii) |x| > R then anx is divergent (iii) |x| = R then all things are possible Proof.

P n (i) As |x| is not sup, there exists an x0 such that |x| < x0 < R and anx0 is convergent. Now use (13.3).

P n (ii) If anx convergent, we would have R > |x| by definition of sup. (iii) Example. P xn/n2 : R = 1 : convergent at 1, −1 P xn/n : R = 1 : convergent at − 1, and divergent at 1 P(−x)n/n : R = 1 : convergent at + 1, and divergent at − 1 P xn : R = 1 : divergent at 1, −1

Note. In the complex case ‘all things are possible’ is not so trivial.

13.6 Examples of working out R

(i) Easy case: If lim an+1/an exists, and = l, then R = 1/l.

n Proof. Put un = |x an|, un+1/un = |x|an+1/an → l|x|. If l|x| < 1 then convergent, if l|x| > 1 then divergent. So by the previous theorem, R = 1/l.

P pn P pn (ii) Comparison Test Case: x , p =sequence of primes. By Comparison test, |x| 6 P |x|n convergent if |x| < 1. But P 1pn is divergent as there are infinitely many primes. (iii) Cosine/sine etc. P n x2n Consider (−1) 2n! : note a0 = 1, a1 = 0, a2 = 1/2etc. So Ratio test is not good - yet! n 2n x2 Put un := |(−1) x /2n!|. Then un+1/un = (2n+2)(2n+1) → 0 as n → ∞. So convergent by Ratio Test! So series absolutely convergent for all x, so convergent for all n. Therefore R = ∞.

2 13.7 Two Big Theorems P n Theorem. Suppose anx has radius of convergence R. Then

P n (i) radius of convergence of (n + 1)an+1x is R

P an n (ii) radius of convergence of n+1 x is R and if |x| < R then

d X  X a xn = (n + 1)a xn dx n n+1 Z x X n X an n+1 ant = x 0 n + 1

Note. There are definitions not content-free. We are not just differentiating/integrating a sum—but a limit!

[Clearly we can’t prove these until we develop the theory of differentiation and integration in HT and TT.]

3 14 The Elementary Functions

Again we will stick to the real case in this section.

14.1 The exponential function (i) For all x, P xn/n! is convergent: so R = ∞.

(ii) Define X xn exp(x) := n! (iii) exp(0) = 1

(iv) exp(x + y) = exp(x) exp(y) [proved]

2 (v) exp(x) = exp(x/2) > 0. (vi) exp(x) > 0 actually. For x > 0, exp(x) > 1 + x > 0 for x = −t, t > 0, exp(x) exp(−x) = exp(0) = 1 So exp(−x) > 0 too.

(vii) xα exp(−x) → 0 as x → ∞.[whatever this means. . . ]

Proof. Chose x ∈ N, n > α. Then for x > 1 exp(x) exp(x) xn+1/(n + 1)! x = xα > xn > xn (n + 1)! So (n + 1)! 0 xα exp(−x) 6 6 x A sanwich argument completes the proof.

(viii) For you: exp(x) = exp(1)x.

[Write x = sup xn, xn ∈ Q, and prove that the result is true for each xn—essentially algebra, done by MI. Passing to the limit is tougher.]

14.2 The Trigonometric Functions (i) For all x X x2n(−1)n X x2n+1(−1)n−1 and 2n! (2n + 1)! are convergent.

(ii) Define X x2n(−1)n X x2n+1(−1)n−1 cos x := , sin x := 2n! (2n + 1)!

4 (iii) cos 0 = 1, sin 0 = 0

(iv) cos x = cos(−x), sin x = − sin(−x)

(v) sin(x + y) = sin x cos y + cos x sin y xrys We identify coefficients of (r+s)! on both sides. Ditto for cos(x + y).

(vi) cos2 x + sin2 x = 1 Well, the first coefficient of each side is 1. Then, clearly only even coefficients can be non-zero. The coefficient of x2k is

in cos2 x: X x2m x2k−2m (−1)m (−1)k−m (2m)! (2k − 2m)! k 1 X  2k  = (−1)k (2k)! 2m m=0 1 = (−1)k22k−1 (2k)!

in sin2 x: X 1 1 (−1)m (−1)k−m−1 (2m + 1)! (2k − 2m − 1)! k−1 (−1)k X  2k  = (−1) (2k)! 2m + 1 m=0 (−1)k = −(−1) 22k−1, (2k)!

So result will follow from what we’ve proved about multiplication of Absolutely con- vergent series.

(vii) Much better:

cos0(x) = − sin x sin0(x) = cos x

[Depends of results we have not proved.]

(viii) Define π/2 := inf{x > 0 : cos(x) = 0}

(ix) Now establish the periodicity:

cos(x + 2π) = cos(x) and sin(x + 2π) = sin(x)

5 14.3 Hyperbolic Functions Define X x2n X x2n+1 cosh x := , sinh x := 2n! (2n + 1)! Then exp(x) + exp(−x) exp(x) − exp(−x) cosh x = , sinh x = 2 2 and all follows.

14.4 The other trigonometric functions Define 1 sec(x) := when x 6= 0 ... cos x sin x tan(x) := ... cos x Next term!

14.5 Logarithm P xn+1 Consider n+1 . The radius of convergence is 1. For |x| < 1 define

X xn+1 log(1 − x) := n + 1

14.6 Binomial Series The series X α(α − 1) ... (α − k + 1) k = 0∞ xk k! has radius of convergence 1 (unless α ∈ N); its sum, when |x| < 1, is

(1 + x)α.

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